Answer:
40, 5 and 3
See explaination for the details
Explanation:
A)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =4bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/4
= 160/4
= 40
b)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =32bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/32
= 160/32
= 5
c)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =64bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/64
= 160/64
= 3
Answer:
1-2
2-3
3-1
Explanation:
I couldn't find a way to explain this.
Answer:
code reviewer
Explanation:
In this scenario, the project manager should advertise for a code reviewer. Usually a development team has at least 2 code reviewers that are very familiar with the code that is being written for the project. Once sections of code have been completed it is usually the case that both the project lead and a code reviewer both review the entire software to look for bugs and note improvements that can be made to make the code more efficient.
Overflow occurs when the magnitude of a number exceeds the range allowed by the size of the bit field. The sum of two identically-signed numbers may very well exceed the range of the bit field of those two numbers, and so in this case overflow is a possibility.
